Uskoro imam pismeni zadatak i hteo bih da proverim sebe, radi se o realnim funkcijama.
Ja cu pokazati kako ja radim, a vi proverite, ako imate savete i upustva podelite ih
[dispmath]y=\frac{x^2-5x+7}{x-2}[/dispmath]
[inlmath]1.[/inlmath] Odrediti domen funkcije
[dispmath]x-2\ne0\\
x\ne2[/dispmath]
[inlmath]2.[/inlmath] Ispitati (ne)parnost funckije
[dispmath]f\left(x\right)=\frac{x^2-5x+7}{x-2}\\
f\left(-x\right)=\frac{\left(-x\right)^2-5\left(-x\right)+7}{\left(-x\right)-2}=\frac{x^2+5x+7}{-x-2}\;\Rightarrow\;f\left(x\right)\ne f\left(-x\right)\\
-f\left(x\right)=-\frac{x^2-5x+7}{x-2}=\frac{-x^2+5x-7}{x-2}\;\Rightarrow\;-f\left(x\right)\ne f\left(-x\right)[/dispmath]
[inlmath]\Rightarrow[/inlmath] funkcija nije ni parna ni neparna.
[inlmath]3.[/inlmath] Asimptote
Vertikalna asimptota
[dispmath]\lim_{x\to2_+}\frac{x^2-5x+7}{x-2}=\frac{1}{0_+}=+\infty[/dispmath]
[inlmath]\Rightarrow\;x=2[/inlmath] je vertikalna asimptota.
Horizontalna asimptota
[dispmath]\lim_{x\to\infty}\frac{x^2-5x+7}{x-2}=\lim_{x\to+\infty}\frac{x-5+\frac{7}{x}}{1-\frac{2}{x}}=\frac{\infty}{1}=\infty[/dispmath]
[inlmath]\Rightarrow[/inlmath] nema horizontalnih asimptota.
Kosa asimptota
[dispmath]k=\lim_{x\to\infty}\frac{\frac{x^2-5x+7}{x-2}}{x}=\lim_{x\to\infty}\frac{x^2-5x+7}{x^2-2x}=\lim_{x\to\infty}\frac{1-\frac{5}{x}+\frac{7}{x^2}}{1-\frac{2}{x}}=\frac{1}{1}=1\\
\Rightarrow\;k=1\\
n=\lim_{x\to\infty}\left[\frac{x^2-5x+7}{x-2}-x\right]=\lim_{x\to\infty}\left[\frac{x^2-5x+7-x\left(x-2\right)}{x-2}\right]=\\
=\lim_{x\to\infty}\frac{x^2-5x+7-x^2+2x}{x-2}=\lim_{x\to\infty}\frac{-3x+7}{x-2}=\lim_{x\to\infty}\frac{-3+\frac{7}{x}}{1-\frac{2}{x}}=-3\\
\Rightarrow\;n=-3[/dispmath]
[inlmath]\Rightarrow\;y=x-3[/inlmath] je kosa asimptota.
[inlmath]4.[/inlmath] Granicne vrednosti
[dispmath]\lim_{x\to+\infty}\frac{x^2-5x+7}{x-2}=\lim_{x\to+\infty}\frac{1-\frac{5}{x}+\frac{7}{x^2}}{\frac{1}{x}-\frac{2}{x^2}}=\frac{1}{0_+}=+\infty\\
\lim_{x\to-\infty}\frac{x^2-5x+7}{x-2}=\lim_{x\to-\infty}\frac{1-\frac{5}{x}+\frac{7}{x^2}}{\frac{1}{x}-\frac{2}{x^2}}=\frac{1}{0_-}=-\infty\\
\lim_{x\to2_+}\frac{x^2-5x+7}{x-2}=\frac{1}{0_+}=+\infty\\
\lim_{x\to2_-}\frac{x^2-5x+7}{x-2}=\frac{1}{0_-}=-\infty[/dispmath]
[inlmath]5.[/inlmath] Prvi izvod
[dispmath]y=\frac{x^2-5x+7}{x-2}\;\Rightarrow\;y'=\frac{\left(x^2-5x+7\right)'\left(x-2\right)-\left(x^2-5x+7\right)\left(x-2\right)'}{\left(x-2\right)^2}=\\
=\frac{\left(2x-5\right)\left(x-2\right)-\left(x^2-5x+7\right)}{\left(x-2\right)^2}=\frac{2x^2-4x-5x+10-x^2+5x-7}{\left(x-2\right)^2}=\frac{x^2-4x+3}{\left(x-2\right)^2}\\
y'=\frac{x^2-4x+3}{\left(x-2\right)^2}[/dispmath]
[inlmath]6.[/inlmath] Stacionarne tacke
[dispmath]y'=0\;\Rightarrow\;\frac{x^2-4x+3}{\left(x-2\right)^2}=0\\
\Rightarrow\;\left(x-2\right)^2\ne0\\
\Rightarrow\;x^2-4x+3=0\\
D=4\\
x_1=3\\
x_2=1[/dispmath]
[inlmath]7.[/inlmath] Ekstremne vrednosti
[dispmath]y_1=f\left(x_1\right)=\frac{3^2-15+7}{3-2}=\frac{1}{1}=1\\
\Rightarrow\;y_1=1\\
y_2=f\left(x_2\right)=\frac{1-5+7}{1-2}=-\frac{3}{1}=-3\\
\Rightarrow\;y_2=-3\\
f\left(x\right)\nearrow x\in\left(-\infty,1\right)\cup\left(3,+\infty\right)\\
f\left(x\right)\searrow x\in\left(1,3\right)\\
\max\left(1,-3\right)\\
\min\left(3,1\right)[/dispmath]
[inlmath]8.[/inlmath] Drugi izvod
[dispmath]y'=\frac{x^2-4x+3}{\left(x-2\right)^2}\;\Rightarrow\;y''=\frac{\left(x^2-4x+3\right)'\left(x-2\right)^2-\left(x^2-4x+3\right)\left(\left(x-2\right)^2\right)'}{\left(x-2\right)^4}=\\
=\frac{\left(2x-4\right)\left(x-2\right)^2-\left(x^2-4x+3\right)\left(2x-4\right)}{\left(x-2\right)^4}=\frac{\left(2x-4\right)\left[\left(x-2\right)^2-\left(x^2-4x+3\right)\right]}{\left(x-2\right)^4}=\\
=\frac{\left(2x-4\right)\left(x^2-4x+4-x^2+4x-3\right)}{\left(x-2\right)^4}=\frac{2x- 4}{\left(x-2\right)^4}=\frac{2\left(x-2\right)}{\left(x-2\right)^4}=\frac{2}{\left(x-2\right)^3}\\
\Rightarrow\;y''=\frac{2}{\left(x-2\right)^3}[/dispmath]
[inlmath]9.[/inlmath] Prevojne tacke
[dispmath]y''=0\;\Rightarrow\;\frac{2}{\left(x-2\right)^3}=0\\
\Rightarrow\;\left(x-2\right)^3\ne0\;\Rightarrow\;\frac{2}{\left(x-2\right)^3}\ne0[/dispmath]
[inlmath]\Rightarrow\;y''\ne0[/inlmath] Nema prevojnih tacaka.
[inlmath]10.[/inlmath] Nule funkcije
[dispmath]y=\frac{x^2-5x+7}{x-2}\\
x^2-5x+7=0\\
D=-3\\
x_{1,2}\notin\mathbb{R}[/dispmath]
Nema realnih nula.
Grafik je crtan rucno u ilustratoru